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(C) The velocity of the sphere before impact is $\vec{v}_1 = 3 \hat{i} + \hat{j}$.The wall is parallel to the $\hat{j}$ axis,meaning the wall lies in

Vedclass · Accepted Answer

$-\hat{i} + \hat{j}$

Vedclass · Answer

(C) The velocity of the sphere before impact is $\vec{v}_1 = 3 \hat{i} + \hat{j}$.The wall is parallel to the $\hat{j}$ axis,meaning the wall lies in the $yz$-plane (normal to the $x$-axis).The component of velocity parallel to the wall ($\hat{j}$ component) remains unchanged during the collision because the impulse acts only along the normal to the wall.Thus,$v_{y, 	ext{after}} = v_{y, 	ext{before}} = 1 \hat{j}$.The component of velocity perpendicular to the wall ($\hat{i}$ component) changes according to the coefficient of restitution $e$.The velocity of approach along the normal is $u_x = 3$.The velocity of separation along the normal is $v_x = -e \cdot u_x = -\frac{1}{3} \cdot 3 = -1$.Therefore,the velocity vector after the collision is $\vec{v}_2 = -1 \hat{i} + 1 \hat{j} = -\hat{i} + \hat{j}$.

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